Compute $\int_{|z|=R} \frac{|dz|}{|z-a|^4}$ when $R>0, |a|\neq R$

Question

I want to compute $\int_{|z|=R} \frac{|dz|}{|z-a|^4}$ when $R>0, |a|\neq R$. I did so by parametrizing the circle, but this way required many pages of tedious calculations.

Do you have a more straightforward way to do it?

Let me mention that my issue here is that we integrate by $|dz|$ not by $dz$, so all the known theorems I know fail to be applied -at least in this form of the integral.

Thanks

Answer 1

Try This

$f^n(z)$ = $\frac {n!}{2(pi)n}$ $\int$ $\frac {f(w) dw}{(w-z)^{n+1}}$

Answer 2

After parametrising $z= R e^{i\theta}$ and after a lot of simplifications you will get $$I = \int_{0}^{2\pi} d\theta \frac{R}{[(R\cos\theta-Re(a))^2+(R\sin\theta-Im(a))^2]^2_{}}.$$ To proceed further : go back to complex variable $Z = e^{i\theta}$, rewrite it in terms of $\cos\theta=\frac{1}{2}(z+\frac{1}{z})$ and $\sin\theta=\frac{1}{2i}(z-\frac{1}{z})$. The resulting integral can be evaluated using complex integration methods.